My claim is that the color breakdown for M&M's follows the advertised distribution. Since the p-value is .289 and the significance level is 0.01, we can conclude that the null hypothesis is failed to reject, there is not sufficient evidence to warrant rejection of the claim that the color breakdown for M&M's follows the advertised distribution (Table B). A bar graph shows that the packet of M&M's had a higher frequency blue, green and orange M&Ms and the least amount of yellow and brown M&Ms (Graph A). It was reasonable for goodness of fit test in terms of the expected counts because we assuming that the color of M&M packet and all expected frequencies were at least 5.

Observed and Expected Values Color Distribution, Test statistics, and bar graph attached

Critical Thinking Question:

1. Do you think selecting a bag of M&Ms is a representative sample of all M&Ms?Explain!
2. Why do you think that the M&M company does not make an equal number of each color of M&Ms?

Table A: Table of Observed and Expected Values to Test M&M's Advertised W Color Observed N Exgected N Residual Blue 30 28.3 1.7 Brown 12 15.3 -3.3 Green 25 23.6 1.4 Orange 26 18.9 7.1 Red 15 15.3 -.3 Yellow 10 16.5 -6.5 Total 1 18 Count Graph A: Bar Graph Showing Color Distribution Orange Color Cases weighted by Count Table B: Table of the Chi-Squared Goodness of Fit Test Test Statistics Color Chi-Square 6.176a df 5 Asymp Sig. .289 a. 0 cells (0.0%) have expected frequencies less than 5. The minimum expected cell frequency is 15.3